Question

In the reaction below, 7.0 mol of NO and 5.0 mol of O₂ are reacted together. The reaction generates 2.3 mol of NO₂. What is the percent yield for the reaction?2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

Percent yield for the reaction = 32.86%

Step-by-step explanation

Given:

Moles of NO = 7.0 mol

Moles of O₂ = 5.0 mol

Moles of NO₂ generated = 2.3mol

2 NO (g) + O₂ (g) → 2 NO₂ (g)

What to find:

The percent yield for the reaction.

Step-by-step solution:

The percent yield for the reaction can be calculated using:

`Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}*100\%`

The actual yield = moles of NO₂ generated = 2.3mol

The theoretical yield can be calculated as follows:

First, we need to determine the limiting reactant, using balanced equation for the reaction.

From the equation, 2 mol NO reacts with 1 mol O₂

So 7.0 mol NO will require (7.0 mol x 1 mol)/2 mol = 3.5 mol O₂

This means the 7.0 mol of NO will be completely consumed by only 3.5mol of O₂. Hence, NO is the limiting reactant and O₂ is the reactant in excess.

The next step is to use the limiting reactant to calculate the theoretical yield for the reaction.

From the reaction equation, 2 mol NO generates 2 mol NO₂

So 7.0 mol NO will generate:

`\frac{7.0\text{ }mol\text{ }NO*2\text{ }mol\text{ }NO_2}{2\text{ }mol\text{ }NO}=7.0\text{ }mol\text{ }NO_2`

Finally, put actual yield = 2.3 mol, and theoretical yield = 7.0 mol into the percent yield formula above:

`\begin{gathered} Percent\text{ }yield=(2.3mol)/(7.0mol)*100\% \\ \\ Percent\text{ }yield=0.3286*100\% \\ \\ Percent\text{ }yield=32.86\% \end{gathered}`